▶️ Answer/Explanation
Solution
(a)
\(\lim_{x \to -\infty} 2xe^{2x} = 0\) (exponential decay dominates linear growth).
\(\lim_{x \to \infty} 2xe^{2x} = \infty\) (exponential growth dominates linear growth).
(b)
First, find the derivative:
\( f'(x) = 2e^{2x} + 4xe^{2x} = 2e^{2x}(1 + 2x) \).
Set \( f'(x) = 0 \):
\( 2e^{2x}(1 + 2x) = 0 \implies x = -\frac{1}{2} \) (since \( e^{2x} \) is never zero).
Evaluate \( f \) at the critical point:
\( f\left(-\frac{1}{2}\right) = 2 \left(-\frac{1}{2}\right) e^{2 \left(-\frac{1}{2}\right)} = -e^{-1} = -\frac{1}{e} \).
This is the absolute minimum because:
- For \( x < -\frac{1}{2} \), \( f'(x) < 0 \) (function is decreasing).
- For \( x > -\frac{1}{2} \), \( f'(x) > 0 \) (function is increasing).
(c)
The range of \( f \) is \( \left[-\frac{1}{e}, \infty\right) \).
(d)
For \( y = bxe^{bx} \), find the derivative:
\( y’ = be^{bx} + b^2xe^{bx} = be^{bx}(1 + bx) \).
Set \( y’ = 0 \):
\( be^{bx}(1 + bx) = 0 \implies x = -\frac{1}{b} \).
Evaluate \( y \) at the critical point:
\( y\left(-\frac{1}{b}\right) = b \left(-\frac{1}{b}\right) e^{b \left(-\frac{1}{b}\right)} = -e^{-1} = -\frac{1}{e} \).
Thus, the absolute minimum value is \(-\frac{1}{e}\) for all nonzero \( b \).